Question

SinA*cosA-?sinA+cosA=0,2​

Answer 1

Ответ:

$\sin( \alpha ) + \cos( \alpha ) = 0.2$

возведем обе части в квадрате

${( \sin( \alpha ) + \cos( \alpha ) )}^{2} = {0.2}^{2} \\ { \sin}^{2} ( \alpha ) + 2 \sin( \alpha ) \cos( \alpha ) + { \cos}^{2} ( \alpha ) = 0.04 \\ 1 + 2 \sin( \alpha ) \cos( \alpha ) = 0.04 \\ \sin( \alpha ) \cos( \alpha ) = \frac{0.04 - 1}{2} \\ \sin( \alpha ) \cos( \alpha ) = \frac{ - 0.96}{2} \\ \sin( \alpha ) \cos( \alpha ) = - 0.48$

Answer 2

Ответ:

(sinA+CosA)*2=0.04

sinA*2+CosA*2+2SinACosA=0.04

2SinACosA=0.96

Sin2A=0.96

Объяснение:

вроде)

Answer 3

It seems you have an equation with trigonometric terms that you'd like to solve for an unknown variable, possibly "?". The equation you've provided is:

Sin(A) * Cos(A) - ? * Sin(A) + Cos(A) = 0.2

To solve this equation for "?", you can factor out the common terms on the left side of the equation:

Sin(A) * Cos(A) - ? * Sin(A) + Cos(A) = 0.2 (sin(A) + cos(A)) * (? - 1) = 0.2

Now, you have an equation in the form of a product:

(sin(A) + cos(A)) * (? - 1) = 0.2

To solve for "?", you can divide both sides of the equation by (sin(A) + cos(A)):

(? - 1) = 0.2 / (sin(A) + cos(A))

Now, add 1 to both sides to solve for "?":

? = 1 + 0.2 / (sin(A) + cos(A))

So, the value of "?" in terms of A is:

? = 1 + 0.2 / (sin(A) + cos(A))

This is the solution for "?", expressed in terms of the variable A and the trigonometric functions sin(A) and cos(A).